Determine the resulting temperature when 150 g of ice at 0 deg.C is mixed with 300 g of water at 50 deg C.
The sum of the heats gained is zero.
150*Hf+ (150+300)c(Tf-50)=0
solve for Tf
To determine the resulting temperature when ice and water are mixed, we can use the principle of conservation of energy. The energy gained by the ice as it heats up is equal to the energy lost by the water as it cools down.
To calculate the energy gained or lost, we can use the specific heat capacity of ice and water, as well as the heat equation:
Q = m * c * ΔT
Where:
Q = amount of heat gained/lost (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in joules/gram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
The specific heat capacity of ice is 2.09 J/g°C, and the specific heat capacity of water is 4.18 J/g°C.
First, let's calculate the energy gained by the ice to reach its final temperature:
Q_ice = m_ice * c_ice * ΔT_ice
Where:
m_ice = mass of ice = 150 g
c_ice = specific heat capacity of ice = 2.09 J/g°C
ΔT_ice = change in temperature of the ice = final temperature - initial temperature = Tf - 0°C
Next, let's calculate the energy lost by the water to reach its final temperature:
Q_water = m_water * c_water * ΔT_water
Where:
m_water = mass of water = 300 g
c_water = specific heat capacity of water = 4.18 J/g°C
ΔT_water = change in temperature of the water = Tf - 50°C
Since energy gained by the ice equals energy lost by the water, we have the equation:
Q_ice = Q_water
Substituting the values, the equation becomes:
m_ice * c_ice * ΔT_ice = m_water * c_water * ΔT_water
Now, let's solve for the final temperature (Tf):
Tf = (m_ice * c_ice * ΔT_ice + m_water * c_water * ΔT_water) / (m_ice * c_ice + m_water * c_water)
Substituting the given values:
Tf = (150 g * 2.09 J/g°C * (Tf - 0°C) + 300 g * 4.18 J/g°C (Tf - 50°C)) / (150 g * 2.09 J/g°C + 300 g * 4.18 J/g°C)
Now, we can solve this equation for Tf using algebraic manipulations.
Please note that this calculation assumes no heat is lost to the surroundings, and the system is isolated. It also assumes the ice completely melts to water.